15 23 23 triangle

Acute isosceles triangle.

Sides: a = 15   b = 23   c = 23

Area: T = 163.0711111789
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 38.06328818502° = 38°3'46″ = 0.66443226111 rad
Angle ∠ B = β = 70.96985590749° = 70°58'7″ = 1.23986350213 rad
Angle ∠ C = γ = 70.96985590749° = 70°58'7″ = 1.23986350213 rad

Height: ha = 21.74328149052
Height: hb = 14.18800966773
Height: hc = 14.18800966773

Median: ma = 21.74328149052
Median: mb = 15.64444878472
Median: mc = 15.64444878472

Inradius: r = 5.34765938291
Circumradius: R = 12.16549382177

Vertex coordinates: A[23; 0] B[0; 0] C[4.89113043478; 14.18800966773]
Centroid: CG[9.29771014493; 4.72766988924]
Coordinates of the circumscribed circle: U[11.5; 3.96768276797]
Coordinates of the inscribed circle: I[7.5; 5.34765938291]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 141.937711815° = 141°56'14″ = 0.66443226111 rad
∠ B' = β' = 109.0311440925° = 109°1'53″ = 1.23986350213 rad
∠ C' = γ' = 109.0311440925° = 109°1'53″ = 1.23986350213 rad

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How did we calculate this triangle?

Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.

a = 15 ; ; b = 23 ; ; c = 23 ; ;

1. The triangle circumference is the sum of the lengths of its three sides

p = a+b+c = 15+23+23 = 61 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 61 }{ 2 } = 30.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30.5 * (30.5-15)(30.5-23)(30.5-23) } ; ; T = sqrt{ 26592.19 } = 163.07 ; ;

4. Calculate the heights of the triangle from its area.

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 163.07 }{ 15 } = 21.74 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 163.07 }{ 23 } = 14.18 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 163.07 }{ 23 } = 14.18 ; ;

5. Calculation of the inner angles of the triangle using a Law of Cosines

a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 15**2-23**2-23**2 }{ 2 * 23 * 23 } ) = 38° 3'46" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-15**2-23**2 }{ 2 * 15 * 23 } ) = 70° 58'7" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-15**2-23**2 }{ 2 * 23 * 15 } ) = 70° 58'7" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 163.07 }{ 30.5 } = 5.35 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 38° 3'46" } = 12.16 ; ;




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